Taking the derivative of e^3x

[sonofjohn]

If f(x) is the function given by f(x) = e3x + 1, at what value of x is the slope of the tangent line to f(x) equal to 2?

I thought the derivative of e3x would be 3e3x because of the chain rule, but it doesn't appear to be correct. I know that ex is just ex, so is e3x just e3x?

[Dick]

You were right the first time. You need the chain rule. (e^(3x))'=3*e^(3x).

[tiny-tim]

Hi sonofjohn! :smile:
I thought the derivative of e3x would be 3e3x because of the chain rule, but it doesn't appear to be correct.

It is correct … who says it isn't? :confused:

[sonofjohn]

Ahh thank you very much!

[sonofjohn]

The region bounded by y = e^x, y = 1, and x = 2 s rotated about the x-axis. The volume of the solid generated is given by the integral_____________.

So this is volume problem and a disc/ washer method formula should work best. I was going to set the problem as:

pi(antider)from 0-2 of (e^x-1)^2(1-1)^2 but I don't believe I will come out with 1-1 and 0 as the volume.

[Dick]

You are rotating around the x-axis. At a given value of x what is the outer and inner radius of the washer?

[sonofjohn]

0 and 2, so they would be my bounds so would I rather set the problem up as, pi*antiderv*(e^x-1)^2

[tiny-tim]

Hi sonofjohn! :smile:
The region bounded by y = e^x, y = 1, and x = 2 s rotated about the x-axis. The volume of the solid generated is given by the integral_____________.
…
pi(antider)from 0-2 of (e^x-1)^2(1-1)^2 but I don't believe I will come out with 1-1 and 0 as the volume.

(have a pi: π :smile:)

I don't actually understand all of your (e^x-1)^2(1-1)^2 …

but you have a π(ex - 1)2, which is not the area of anything, is it? :wink:

[Dick]

No, at say x=1 what is the inner and outer radius of the washer? What about at a general value of x? 0 and 2 are fine for the limits on the x integration. Now you just want to find the area of the washer. Correctly, this time. It's not pi*(e^x-1)^2.

[sonofjohn]

Ok so I should find the area of the washer. The area of the washer should be defined as e^x - 1 Also since I am using the dish washer formula, I should square both parts of the integration thus yielding:

pi(anitderivative)(e^2x -1)

[Dick]

Ok. Yes. The area is pi*(outer radius^2-inner radius^2).

[sonofjohn]

Ok. Yes. The area is pi*(outer radius^2-inner radius^2).

Thanks for the clarification!

[sonofjohn]

http://i203.photobucket.com/albums/aa40/cjohnson1992/IM000000.jpg

The 3rd problem on this page is difficult for me to understand. I don't understand what the x stands for when they are talking about the bounds in terms of integration. Do they mean the x-axis on the graph or possible x = 2 where at the x intercept?

[Dick]

x is a point between 0 and 4. Any one. What you might notice is that G(x) is the area under the curve from 0 to x. H(x) is NEGATIVE of the area under the curve from x to 2 (because the integral is from 2 to x instead of from x to 2). Might this tell you something about G(x)-H(x)?

[sonofjohn]

So then the answer must be g(x) = h(x) - 2, because h(x) is positive and g(x) is negative. g(x) is also always going to be two less, because it goes from any point to 0, and h(x) only goes to 2.

[Dick]

So then the answer must be g(x) = h(x) - 2, because h(x) is positive and g(x) is negative. g(x) is also always going to be two less, because it goes from any point to 0, and h(x) only goes to 2.

I can't really agree with you there, sonofjohn. Try this. What are G(1) and H(1)? Work them out from the picture.

[sonofjohn]

I see now that (d) cannot work. Subtracting 2 everytime from h will not yield an equal integral. Now I would like to say that g(x) = h(x+2) would work, but it doesn't seem plausible past h(1). Could G'(x) = H'(x+2) work? I don't even understand what it means.

[Dick]

You didn't answer my last question. What are G(1) and H(1)? That should let you eliminate some possibilities.

[sonofjohn]

g(1) is -2 and h(1) is 1 so a,d,e cannot be the right answer .

[Dick]

g(1) is -2 and h(1) is 1 so a,d,e cannot be the right answer .

How did you get that? I get G(1)=2 and H(1)=(-1).

[sonofjohn]

Well, wouldn't you be going from 1-0 with g and therefore getting a negative integral and then 1-2 with h and getting a positive integral?

[Dick]

Well, wouldn't you be going from 1-0 with g and therefore getting a negative integral and then 1-2 with h and getting a positive integral?

It sure looks to me like G is the integral from 0 to x, not x to 0. I've checked the problem several times.

[sonofjohn]

Alright I see that, had the bounds mixed up :( So if G(1) is 2 and h(1) is -1, does that eliminate any other possibilities?

[Dick]

Use it to check a) d) and e). Now what is G(0) and H(0)? Or G(2) and H(2)?

[sonofjohn]

Ok, g(0) = 0 and h(0) = -3
g(2) = 3 and h(2) = 0

They seem to be opposite of eachother.

[Dick]

G(1)=2 and H(1)=(-1). Those aren't opposites, just different sign. Don't you see the relation between the two yet?

[sonofjohn]

Yes, I think so. If I were to add 3 to the h(x) or (e) I should always get g(x).

[Dick]

Sure. G(x) is the area between 0 and x. H(x) is the NEGATIVE of the area between x and 2. So G(x)-H(x)=(area between 0 and 2)=3.

[sonofjohn]

Ahh very good to know!

[Mark44]

sonofjohn,
One more comment from the peanut gallery...

If the problem had been stated like this:
F(x) = \int_0^x{f(t)dt}
you would get into big trouble if you used f when you meant F or vice versa. If you'll notice Dick's comments, he was always very careful to use exactly the same letters (including capitalization) as were given in the problem.

[sonofjohn]

I see, I'll make sure to keep the numbers and letters the same as they were in the problem.